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# Terrence writes down one of the numbers from 0-20 on one index card ea

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Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 01:21
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Difficulty:

85% (hard)

Question Stats:

49% (03:07) correct 51% (03:25) wrong based on 88 sessions

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Terrence writes down one of the numbers from 0-20, inclusive, on one index card each until he has written each number exactly once and then faces all the cards down. Next, he randomly chooses two cards without turning them over. What is the probability that a prime number will be written on each card and that the absolute difference between the two prime numbers will itself be a prime number?

(A) 1/35

(B) 4/95

(C) 4/105

(D) 8/95

(E) 8/105

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Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 03:53
The total ways of picking up 2 numbers from 21 numbers are $$21c2 = \frac{21*20}{2}= 210$$

Since we have been asked to find out the probability of finding 2 prime numbers
whose difference is a prime number, there are 8 such possibilities
For the number pairs {(2,5),(3,5),(2,7),(5,7),(2,13),(11,13),(2,19),(17,19)}

The probability that a prime number will be written on each card and that
the absolute difference between the two prime numbers will also be a prime number is $$\frac{8}{210}$$ = $$\frac{4}{105}$$(Option C)
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Re: Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 05:01
Ans :C

21C2=210
And possibilities are 8 (3,5)(5,2)(5,7)(7,2)(11,13)(13,2)(17,19)(19,2)
8/210=4/105

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Re: Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 05:36
Bunuel wrote:
Terrence writes down one of the numbers from 0-20, inclusive, on one index card each until he has written each number exactly once and then faces all the cards down. Next, he randomly chooses two cards without turning them over. What is the probability that a prime number will be written on each card and that the absolute difference between the two prime numbers will itself be a prime number?

(A) 1/35

(B) 4/95

(C) 4/105

(D) 8/95

(E) 8/105

Total number of cards are 21
21C2 Card choice
Denominator will be (21*20)/2 = 210 Option D and B are out

Prime number 2, 3, 5, 7, 11, 13, 17, 19
Difference of prime numbers be prime (Favorable selection) = (2,5) (2,7), (2,13), (2,19), (3,5), (5,7), (11,13), (17, 19) and also their reversed order i.e. (5,2) (7,2).....

(8*2)/210

8/105

Option E
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Re: Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 05:50
1
Bunuel wrote:
Terrence writes down one of the numbers from 0-20, inclusive, on one index card each until he has written each number exactly once and then faces all the cards down. Next, he randomly chooses two cards without turning them over. What is the probability that a prime number will be written on each card and that the absolute difference between the two prime numbers will itself be a prime number?

(A) 1/35

(B) 4/95

(C) 4/105

(D) 8/95

(E) 8/105

Prime numbers from 0 to 20 = 2, 3, 5, 7, 11, 13, 17, 19 = 8 values.

Let $$a$$ be the first card and $$b$$ be the second card chosen.

Probability that each card has a prime number written on its face = $$\frac{8}{21} * \frac{7}{20} = \frac{2}{3} * \frac{1}{5} = \frac{2}{15}$$.

Let $$(a,b)$$ be a pair such that $$|a-b|$$ is a prime number, such pairs are $$(2,5), (2,7), (2,13), (2,19), (3,5), (5,7), (11,13), (17,19)$$.

No. of ways 2 cards can be chosen from 8 = $$8c2 = 28$$.

Probability that $$|a-b|$$ is a prime number = $$8/28 = 2/7$$.

Required probability = $$\frac{2}{15} * \frac{2}{7} = \frac{4}{105}$$. Ans - C.
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Re: Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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27 Jul 2017, 09:11
mynamegoeson wrote:
Bunuel wrote:
Terrence writes down one of the numbers from 0-20, inclusive, on one index card each until he has written each number exactly once and then faces all the cards down. Next, he randomly chooses two cards without turning them over. What is the probability that a prime number will be written on each card and that the absolute difference between the two prime numbers will itself be a prime number?

(A) 1/35

(B) 4/95

(C) 4/105

(D) 8/95

(E) 8/105

Total number of cards are 21
21C2 Card choice
Denominator will be (21*20)/2 = 210 Option D and B are out

Prime number 2, 3, 5, 7, 11, 13, 17, 19
Difference of prime numbers be prime (Favorable selection) = (2,5) (2,7), (2,13), (2,19), (3,5), (5,7), (11,13), (17, 19) and also their reversed order i.e. (5,2) (7,2).....

(8*2)/210

8/105

Option E

Hi mynamegoeson,
Since absolute difference between primes is required, does it matter whether the guy picks up in the sequence of 11-13 or 13-11?
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Re: Terrence writes down one of the numbers from 0-20 on one index card ea  [#permalink]

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28 Jul 2017, 04:19
Dkingdom wrote:
mynamegoeson wrote:
Bunuel wrote:
Terrence writes down one of the numbers from 0-20, inclusive, on one index card each until he has written each number exactly once and then faces all the cards down. Next, he randomly chooses two cards without turning them over. What is the probability that a prime number will be written on each card and that the absolute difference between the two prime numbers will itself be a prime number?

(A) 1/35

(B) 4/95

(C) 4/105

(D) 8/95

(E) 8/105

Total number of cards are 21
21C2 Card choice
Denominator will be (21*20)/2 = 210 Option D and B are out

Prime number 2, 3, 5, 7, 11, 13, 17, 19
Difference of prime numbers be prime (Favorable selection) = (2,5) (2,7), (2,13), (2,19), (3,5), (5,7), (11,13), (17, 19) and also their reversed order i.e. (5,2) (7,2).....

(8*2)/210

8/105

Option E

Hi mynamegoeson,
Since absolute difference between primes is required, does it matter whether the guy picks up in the sequence of 11-13 or 13-11?

Yes i think it should be 4/105 since difference is required
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Re: Terrence writes down one of the numbers from 0-20 on one index card ea   [#permalink] 28 Jul 2017, 04:19
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